Physics
posted by Alphonse on .
A 19.5 gram, 56.3 centimeter long steel guitar string is vibrating in its fourth harmonic. It induces a standing wave in a nearby 48.6 centimeter metal pipe open at one end. The standing sound wave in the pipe corresponds to the pipe's fundamental frequency.
What is the tension on the steel guitar 'string'? (in Newtons)

Just figured this one out! Okay, so:
The frequency of the pipe(f_p), and of the string (f_s) are both the same, since frequency is source dependent.
f_p=f_s
Now, f_p=n(v_snd/4L_p)
where n is 1 since the wave corresponds to the pipe's fundamental (first) frequency, v_snd is the speed of sound and L_p is the length of the pipe. For you, this value should come to: f_p=176.4403292...Hz, which is also f_s.
f_s=n(v_str/2L_s)
This time, we're looking for v_str (speed that the string vibrates). Rearranging gives us:
v_str=[2(L_s)(f_s)]/n
where n=4 since the string vibrates in its fourth harmonic, so your v_str=49.66795267...m/s
Finally, using the formula
v_str=√[T/(m/L_s)] and rearranging gives you T=m*((v_str)^2)/L_s
So your tension would be: 85.44344174...N
Check my work just in case! 
Oh, btw, did you figure out questions 4,5, or 7?

Sorry for the late reply. Sadly, I didn't any of the ones you mentioned. Did you get #3?

For #4, it's F_tension=density x (lambda/T)^2

For #3, I put B. I dunno if it's the same for everyone's (since it's MC)